quantum mechanics - Griffiths Intro to QM Section 9.1.2: What type of approximation is he using here and what is the justification for it?

I really do not understand Griffiths logic in this section and was wondering if someone could help. This is basically a 1st order coupled system of ordinary differential equations but I haven't seen an approximation like this before.

For brevity $f(t)=-\frac{i}{\hbar}H'_{ab}e^{-i\omega_0t}$ and $g(t)=-\frac{i}{\hbar}H'_{ba}e^{i\omega_0t}$ so the system of equations are (from Griffiths equation 9.13)

$$\begin{align*} \dot c_a &= f(t)\,c_b\\ \dot c_b &= g(t)\,c_a \hspace{10mm} [9.13] \end{align*}$$

Note both $c_a$ and $c_b$ are functions of time. Griffiths then goes on to express the derivative of a $n$th order approximation as proportional to the next lower order approximation of the other system variable.

For example in equation 9.18 Griffiths states

$$\frac{d c_a^{(2)}}{dt} = f(t) \, c_b^{(1)} \hspace{10mm} [9.18]$$

I just don't see how to justify this? (Note: Griffith says his superscript in parentheses indicates the order of the approximation.)

So from my reading of this $c_a^{(n)}$ and $c_b^{(n)}$ are just $n$th order expansions of $c_a$ and $c_b$. So in my mind I am thinking if we are approximating each out to $n$ terms them something like equation 9.18 above should instead be

$$\frac{d c_a^{(2)}}{dt} = f(t) \, c_b^{(2)} \hspace{10mm}$$

In other words we are taking the derivative of $c_a$ and just approximating it out to say 2 terms, then shouldn't we use the same order approximation for $c_b$ in the system of equations [9.13]?

So why can Griffiths do this? Why can we just stick in the lower order approximation to solve for the next one up?

Answer

This is a typical perturbative expansion, although presented in a more pedestrian manner.

What is usually done for convenience of expansion, is to attach to $H'$ a (time-independent) coupling or scale constant, say $H' \rightarrow \lambda H'$, and to make explicit the assumption that solutions are sought as perturbative expansions in $\lambda$: $$c_a(t) = c_a^{(0)}(t) + \lambda c_a^{(1)}(t) + \lambda^2 c_a^{(2)}(t) + \dots\\ c_b(t) = c_b^{(0)}(t) + \lambda c_b^{(1)}(t) + \lambda^2 c_b^{(2)}(t) + \dots$$ When substituted in your eqs.(9.13), these expansions generate a hierarchy of differential equations from the demand that the polynomial expansions hold for arbitrary $\lambda$. That is, first obtain $$\dot c_a^{(0)}(t) + \lambda \dot c_a^{(1)}(t) + \lambda^2 \dot c_a^{(2)}(t) + \dots = \lambda f(t)\left[ c_b^{(0)}(t) + \lambda c_b^{(1)}(t) + \lambda^2 c_b^{(2)}(t) + \dots \right]\\ \dot c_b^{(0)}(t) + \lambda \dot c_b^{(1)}(t) + \lambda^2 \dot c_b^{(2)}(t) + \dots = \lambda g(t)\left[ c_a^{(0)}(t) + \lambda c_a^{(1)}(t) + \lambda^2 c_a^{(2)}(t) + \dots \right]$$ then identify the coefficients of successive powers of $\lambda$:

$\lambda^0$: $$\dot c_a^{(0)}(t) = 0\\ \dot c_b^{(0)}(t) = 0$$ $\lambda$: $$\dot c_a^{(1)} = f(t) c_b^{(0)}\\ \dot c_b^{(1)} = g(t) c_a^{(0)}$$ $\lambda^2$: $$\dot c_a^{(2)} = f(t) c_b^{(1)}\\ \dot c_b^{(2)} = g(t) c_a^{(1)}$$ In general, for $k\ge 1$, $$\dot c_a^{(k)} = f(t) c_b^{(k-1)}\\ \dot c_b^{(k)} = g(t) c_a^{(k-1)}$$ The rest follows from successively solving lower order eqs. and substituting in the next order set.